Problems with Scanf("%D\N",&I)

what happens when you input things like 12ab to scanf(%d,&argu)?

It's better to read a full line, using fgets(), and then inspecting it, rather than trying to parse "on the fly" from the input stream.

It's easier to ignore non-valid input, that way.

Use fgets() and then just strtol() to convert to a number, it will make it easy to see if there is trailing data after the number.

For instance:

char line[128];

while(fgets(line, sizeof line, stdin) != NULL)
{
   char *eptr = NULL;
   long v = strtol(line, &eptr, 10);
   if(eptr == NULL || !isspace(*eptr))
   {
     printf("Invalid input: %s", line);
     continue;
   }
   /* Put desired processing code here. */
}

scanf with %n giving wrong output

%n captures all the characters processed by scanf including leading whitespace. Using %n twice can correct that. The format string skips leading whitespace and then gets the beginning count of characters. Then the integer is scanned and finally the total count of characters is captured. The difference in the count is the characters in the integer.

Always check the return of scanf as the input stream may need cleaning.

    int begin = 0;
    int end = 0;
    int val = 0;
    int clean = 0;
    int result = 0;

    do {
        if ( ( result = scanf(" %n%d%n",&begin,&val,&endn)) != 1) {// scan one int
            while ( ( clean = getchar ( )) != '\n') {//clean bad input
                if ( clean == EOF) {
                    fprintf ( stderr, "problem reading input\n");
                    exit ( 1);
                }
            }
        }
        else {//scanf success
            printf("byte count is: %d\n",end-begin);
        }
    } while ( result != 1);

Scanf does not work as expected

scanf("%d\n",&i); is equivalent to std::cin >> i >> std::ws;.

If you want the same behaviour for scanf, remove \n: scanf("%d",&i);

This is caused by the fact that any whitespace character in scanf means "skip input until non-whitespace is found"

basic problems in scanf format and using function pointer

scanf is a nasty beast. The difference between using scanf("%d %d ", &i, &j); and scanf("%d %d", &i, &j); is the extra whitespace at the end of the former statement. This makes scanf want to skip more whitespace; but because there isn't any, it asks the console to get more (prompting you for input). Once you press ENTER, it takes that as a newline and gobbles that up - then scanf lets you continue.

Also I wonder why input those 2 two numbers in this way scanf("%d %d ", &i, &j); instead of scanf("%d %d ", i, j);?

This is because scanf needs to modify the integers you are passing to it. In C, if you want to modify your argument then you need to get passed the pointer to the value. If the value itself is parsed and taken as a pointer, then you'd most likely get a segmentation fault.

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